Current Electricity Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 24 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, each \( 4 \, \Omega \). What is the total current?

Equivalent resistance: \( R_{\text{eq}} = \frac{5}{6} R = \frac{5}{6} \times 4 = \frac{20}{6} = \frac{10}{3} \approx 3.33 \, \Omega \).

Total current: \( I = \frac{V}{R_{\text{eq}}} = \frac{24}{\frac{10}{3}} = 24 \times \frac{3}{10} = 7.2 \, \text{A} \).

\( 6.0 \, \text{A} \)
\( 6.5 \, \text{A} \)
\( 7.2 \, \text{A} \)
\( 8.0 \, \text{A} \)
3

A \( 20 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 2.5 \, \text{A} \) to a resistor. What is the resistance of the resistor?

Terminal voltage: \( V = \varepsilon - I r = 20 - 2.5 \times 2 = 15 \, \text{V} \).

Resistance: \( R = \frac{V}{I} = \frac{15}{2.5} = 6 \, \Omega \).

\( 5.0 \, \Omega \)
\( 5.5 \, \Omega \)
\( 6.0 \, \Omega \)
\( 6.5 \, \Omega \)
3

A \( 4 \, \Omega \) resistor dissipates \( 16 \, \text{W} \) of power. What is the voltage across it?

Power: \( P = \frac{V^2}{R} \).

Rearrange: \( V = \sqrt{P R} \).

Substitute: \( V = \sqrt{16 \times 4} = \sqrt{64} = 8 \, \text{V} \).

\( 8 \, \text{V} \)
\( 10 \, \text{V} \)
\( 12 \, \text{V} \)
\( 16 \, \text{V} \)
1

A copper wire carries \( 3 \, \text{A} \) with a drift speed of \( 9 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), what is the cross-sectional area?

Drift speed: \( v_d = \frac{I}{n e A} \).

Rearrange: \( A = \frac{I}{n e v_d} \).

Substitute: \( A = \frac{3}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 9 \times 10^{-5}} \).

Calculate: \( A = \frac{3}{1.224 \times 10^5} \approx 2.45 \times 10^{-5} \, \text{m}^2 \).

\( 2.0 \times 10^{-5} \, \text{m}^2 \)
\( 2.2 \times 10^{-5} \, \text{m}^2 \)
\( 2.4 \times 10^{-5} \, \text{m}^2 \)
\( 2.45 \times 10^{-5} \, \text{m}^2 \)
4

A \( 15 \, \text{V} \) battery with negligible internal resistance is connected to a \( 3 \, \Omega \) and \( 12 \, \Omega \) resistor in series. What is the power dissipated in the \( 12 \, \Omega \) resistor?

Total resistance: \( R = 3 + 12 = 15 \, \Omega \).

Current: \( I = \frac{V}{R} = \frac{15}{15} = 1 \, \text{A} \).

Power: \( P = I^2 R = 1^2 \times 12 = 12 \, \text{W} \).

\( 10 \, \text{W} \)
\( 12 \, \text{W} \)
\( 14 \, \text{W} \)
\( 16 \, \text{W} \)
2

A \( 8 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 7 \, \Omega \) resistor. What is the power dissipated in the internal resistance?

Total resistance: \( R_{\text{total}} = 7 + 1 = 8 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{8}{8} = 1 \, \text{A} \).

Power: \( P = I^2 r = 1^2 \times 1 = 1 \, \text{W} \).

\( 0.5 \, \text{W} \)
\( 0.8 \, \text{W} \)
\( 0.9 \, \text{W} \)
\( 1.0 \, \text{W} \)
4

A \( 12 \, \Omega \) resistor dissipates \( 48 \, \text{W} \) of power. What is the current through it?

Power: \( P = I^2 R \).

Rearrange: \( I = \sqrt{\frac{P}{R}} \).

Substitute: \( I = \sqrt{\frac{48}{12}} = \sqrt{4} = 2 \, \text{A} \).

\( 2.0 \, \text{A} \)
\( 2.5 \, \text{A} \)
\( 3.0 \, \text{A} \)
\( 4.0 \, \text{A} \)
1

A wire of length \( 5 \, \text{m} \) and resistance \( 20 \, \Omega \) is stretched to \( 10 \, \text{m} \). What is the new resistance?

Volume constant: \( l A = l' A' \Rightarrow A' = \frac{A}{2} \).

New resistance: \( R' = \frac{\rho l'}{A'} = \frac{\rho (2l)}{\frac{A}{2}} = 4 \frac{\rho l}{A} = 4R = 4 \times 20 = 80 \, \Omega \).

\( 60 \, \Omega \)
\( 70 \, \Omega \)
\( 80 \, \Omega \)
\( 90 \, \Omega \)
3

A \( 20 \, \text{V} \) battery with negligible internal resistance is connected to a \( 5 \, \Omega \) and \( 15 \, \Omega \) resistor in series. What is the power dissipated in the \( 15 \, \Omega \) resistor?

Total resistance: \( R = 5 + 15 = 20 \, \Omega \).

Current: \( I = \frac{V}{R} = \frac{20}{20} = 1 \, \text{A} \).

Power: \( P = I^2 R = 1^2 \times 15 = 15 \, \text{W} \).

\( 12 \, \text{W} \)
\( 15 \, \text{W} \)
\( 18 \, \text{W} \)
\( 20 \, \text{W} \)
2

A Wheatstone bridge with \( R_1 = 9 \, \Omega \), \( R_2 = 18 \, \Omega \), \( R_3 = 12 \, \Omega \), \( R_4 = 24 \, \Omega \) has a \( 20 \, \text{V} \) battery across AC and a galvanometer (\( 5 \, \Omega \)) across BD. What is the current through the galvanometer?

Check balance: \( \frac{R_1}{R_2} = \frac{9}{18} = 0.5 \), \( \frac{R_3}{R_4} = \frac{12}{24} = 0.5 \). Bridge is balanced.

Since balanced, \( I_g = 0 \, \text{A} \).

\( 0.1 \, \text{A} \)
\( 0.05 \, \text{A} \)
\( 0.02 \, \text{A} \)
\( 0 \, \text{A} \)
4

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